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Let K = $\mathbb{Q}(√2,√3)$ be the splitting field of $(x^2−2)(x^2−3)$ over $\mathbb{Q}$. I know that $K = \mathbb{Q}(\alpha)$ with $\alpha =√2 + √3$, and that the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $f(x) = (x^2 − 5)^2 − 24$

I want to show that the splitting field of $f$ over $\mathbb{Q}$ is $K$ and express the elements of $\Gamma(K : \mathbb{Q})$ as permutations of the set of roots of $f$.

Teddy38
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Bradley Hill
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  • Hint: show that if $f$ splits in $E$ (over $\Bbb Q$), then $K \subseteq E$ (it suffices to shows that $\sqrt{2},\sqrt{3} \in E$). As for finding the Galois group, you can start by realizing you're looking for a subgroup of $S_4$ that doesn't have any $3$-cycles, and that such a permutation of the roots cannot send a square root of $2$ to a square root of $3$, – David Wheeler Nov 07 '17 at 00:20

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Actually the roots of $f$ are $\pm \sqrt{2}\pm\sqrt{3}$, so the splitting field is exactly $K$.

For the Galois group, consider the convolutions $\sigma\in \mathrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}(\sqrt{3})) \colon \sqrt{2}\mapsto -\sqrt{2}$ and $\tau\in \mathrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}(\sqrt{2}))\colon \sqrt{3}\mapsto -\sqrt{3}$. Both of them are elements in $\mathrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$ of order 2. Here is a more explicit expression

$$\begin{array}{c} \sigma\colon \pm\sqrt{2}\pm\sqrt{3}\mapsto \mp\sqrt{2}\pm\sqrt{3}\\ \tau\colon \pm\sqrt{2}\pm\sqrt{3}\mapsto \pm\sqrt{2}\mp\sqrt{3}\\ \sigma\tau=\tau\sigma\colon \pm\sqrt{2}\pm\sqrt{3}\mapsto \mp\sqrt{2}\mp\sqrt{3}\\ \end{array} $$

You can also find that the Galois group $\{1,\sigma,\tau,\sigma\tau\}$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$.

Phil. Z
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    This question has been asked many times under different forms. A quick answer is that your field $K$ is the compositum of the 2 quadratic fields $\mathbf Q(\sqrt 2, \sqrt 3)$, which are distinct (why ?) hence linearly disjoint, so $K/\mathbf Q$ is Galois, with group $\cong (\mathbf Z/2)^2$. It can be shown that $K=\mathbf Q(\sqrt 2 + \sqrt 3)$ (a good exercise). The usual next question is to consider $K= \mathbf Q(\sqrt p_1, ...,\sqrt p_n) $ , where $p_1,...,p_n$ are distinct prime numbers, and show inductively that it is Galois with group $\cong (\mathbf Z/2)^n$ (also a good exercise). – nguyen quang do Nov 07 '17 at 15:25
  • See https://math.stackexchange.com/a/1609061/300700 (modulo MathJax !) – nguyen quang do Nov 08 '17 at 07:53